[References] The specific rotation, [a]p, for sucrose is +67. What is the observed rotation for a solution of 0.50 g of sucrose in 10 mL of water in a sample tube having a pathlength of 10 cm? degrees. The observed rotation of a solution of 1.2 g of a compound in 10 mL of water is +1.3 degrees. If the pathlength is 10 cm, what is the specific rotation of the compound?

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Question 19
**[References]**

The specific rotation, \([α]_D\), for sucrose is \( +67 \). What is the observed rotation for a solution of \( 0.50 \) g of sucrose in \( 10 \) mL of water in a sample tube having a path length of \( 10 \) cm?

\[\_\_\_\_\_\] degrees.

The observed rotation of a solution of \( 1.2 \) g of a compound in \( 10 \) mL of water is \( +1.3 \) degrees. If the path length is \( 10 \) cm, what is the specific rotation of the compound?

\[\_\_\_\_\_\].

---

In this image, there are two questions related to the concept of optical rotation and specific rotation in chemistry. Here’s a detailed explanation for each part of the questions.

### Optical Rotation and Specific Rotation
Optical rotation refers to the rotation of the plane of polarized light by a chiral compound. The degree of this rotation is dependent on the concentration of the compound in solution, the path length through which the light passes, and the specific rotation (\([α]_D\)) of the compound.

#### Formula:
\[ [α]_D = \frac{α}{c \cdot l} \]

Where:
- \([α]_D\) = specific rotation
- \(α\) = observed rotation (in degrees)
- \(c\) = concentration of the solution (in g/mL)
- \(l\) = path length (in dm)

### Explanation of Questions
1. **First Question**:
   - Given:
     - \([α]_D\) for sucrose = \( +67 \)
     - 0.50 g of sucrose in 10 mL of water
     - Path length = 10 cm (or 1 dm)

   - To find: Observed rotation for this solution.

   - Calculation:
     - Convert the concentration from g/10 mL to g/mL: \( \frac{0.50 \text{ g}}{10 \text{ mL}} = 0.05 \text{ g/mL} \)
     - Use the formula \( α = [α]_D \cdot c \cdot l \)
     - Substitute the values: \( α = 67 \times 0.05 \times 1 =
Transcribed Image Text:**[References]** The specific rotation, \([α]_D\), for sucrose is \( +67 \). What is the observed rotation for a solution of \( 0.50 \) g of sucrose in \( 10 \) mL of water in a sample tube having a path length of \( 10 \) cm? \[\_\_\_\_\_\] degrees. The observed rotation of a solution of \( 1.2 \) g of a compound in \( 10 \) mL of water is \( +1.3 \) degrees. If the path length is \( 10 \) cm, what is the specific rotation of the compound? \[\_\_\_\_\_\]. --- In this image, there are two questions related to the concept of optical rotation and specific rotation in chemistry. Here’s a detailed explanation for each part of the questions. ### Optical Rotation and Specific Rotation Optical rotation refers to the rotation of the plane of polarized light by a chiral compound. The degree of this rotation is dependent on the concentration of the compound in solution, the path length through which the light passes, and the specific rotation (\([α]_D\)) of the compound. #### Formula: \[ [α]_D = \frac{α}{c \cdot l} \] Where: - \([α]_D\) = specific rotation - \(α\) = observed rotation (in degrees) - \(c\) = concentration of the solution (in g/mL) - \(l\) = path length (in dm) ### Explanation of Questions 1. **First Question**: - Given: - \([α]_D\) for sucrose = \( +67 \) - 0.50 g of sucrose in 10 mL of water - Path length = 10 cm (or 1 dm) - To find: Observed rotation for this solution. - Calculation: - Convert the concentration from g/10 mL to g/mL: \( \frac{0.50 \text{ g}}{10 \text{ mL}} = 0.05 \text{ g/mL} \) - Use the formula \( α = [α]_D \cdot c \cdot l \) - Substitute the values: \( α = 67 \times 0.05 \times 1 =
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